20 November 2025 · Shift 1

Tool and its function In each pair the first word is an instrument and the second is the action that the instrument is meant to perform. A pen is the instrument used to write. In the same way a knife is the instrument used to cut, so 'Cut' completes the analogy. 'Sharp' is only a property of a knife and 'Steel' is the material it is made of, while 'Food' is merely one of the things a knife may cut — none of these is the action it performs.

Profit and loss Profit is always reckoned on the cost price, so a profit of 20% makes the selling price 120% of the cost price. Formula: $SP=CP\times \frac{100+P}{100}$, where P is the profit per cent. Putting SP = 900 and P = 20: $900=CP\times \frac{120}{100}$ $CP=900\times \frac{100}{120}=750$ So the cost price of the article is ₹750.

Number series (squares) Test the terms for perfect squares: $2^2=4$, $3^2=9$, $4^2=16$, $5^2=25$, $6^2=36$. So the series is simply the squares of the consecutive natural numbers beginning at 2. The next term must therefore be the square of 7, that is $7^2=49$. The same thing is seen in the gaps +5, +7, +9, +11, which grow by 2 each time, so the next gap is +13 and 36+13=49.

Interchange of numbers As it stands the left side gives $6+3\times 2-4=6+6-4=8$, which is not 16, so an interchange is needed. Interchanging 4 and 2 turns the equation into $6+3\times 4-2$. By BODMAS the multiplication is done first: $3\times 4=12$. Then 6+12-2 = 16, which is exactly the required value. The other interchanges fail: $4+3\times 2-6=4$, $6+3-2\times 4=1$ and $6+4\times 2-3=11$.

Organ and its sense Each pair joins a sense organ with the sensation that the organ produces. The hand is the organ through which we feel touch. In the same way the nose is the organ through which we smell, so 'Smell' completes the analogy. Taste belongs to the tongue, sight to the eye and hearing to the ear, so each of the other options names the sense of a different organ.

Number series (cubes) Test the terms for perfect cubes: $5^3=125$, $6^3=216$, $7^3=343$, $8^3=512$, $9^3=729$. So the series is the cubes of consecutive natural numbers beginning at 5. The next term is therefore ${10}^3=1000$. Among the other choices 2197 is ${13}^3$ and 2744 is ${14}^3$, so they come far later in the pattern, while 1029 is not a cube at all.

Letter-number coding In the given pair every letter has simply been replaced by its position in the English alphabet. N is the 14th letter, Q the 17th and B the 2nd, which gives 14-17-2. Now read the numbers 5-15-20 back as letters: the 5th letter is E, the 15th is O and the 20th is T. So the group that stands for 5-15-20 is EOT.

Letter-number series Every term is made of a letter and a number, so the two parts must be examined separately. The letters run backwards from the end of the alphabet: Z, Y, X, W, so the next letter is V. The numbers are the cubes of the counting numbers: $1^3=1$, $2^3=8$, $3^3=27$, $4^3=64$. The fifth number is therefore $5^3=125$, and joining the two parts gives V125.

Address matching Compare each candidate address with the master address one field at a time — house number, sector, locality, block and PIN code. Option (a) changes the PIN code from 700106 to 700166. Option (b) changes the block from AC to DC. Option (d) changes the house number from 42 to 43. Only option (c) reproduces every field, every comma and every space exactly as in the master address, so it is the correct match.

Address matching Check the candidates against the master address field by field — shop number, level, name of the building, landmark and PIN code. Option (a) has Shop 8 in place of Shop 7. Option (c) has Level 3 in place of Level 2. Option (d) has the PIN code 422043 in place of 422003. Option (b) agrees with the master address in every field and in every punctuation mark, so it is the exact match.

Odd one out Group the four body parts by the work each one does. The eye, the ear and the nose are all sense organs — they carry sight, hearing and smell to the brain. The lung is not a sense organ at all; it belongs to the respiratory system and exchanges oxygen and carbon dioxide. So the lung does not fit the group and is the odd one out.

Decision making Weigh only what has actually been observed: the student keeps glancing at his mobile phone, but no act of copying has been seen. The response must therefore be in proportion to the evidence — it should stop the risk at once without punishing an offence that has not been established. Cancelling the paper (option a) is a final punishment and cannot be justified when nothing has been proved. Taking the phone away (option c) or reporting the matter to the management (option d) are penal steps that belong to a later stage, if the student ignores the caution. Warning the student and asking him to be careful puts him on notice, ends the suspicious behaviour immediately and keeps the examination hall calm. Hence option (b) is the correct course of action.

Embedded figure First list the parts of the question figure: a curved top edge, a straight base well below it, and two slanting lines that join the ends of the curve to the opposite ends of the base and cross each other on the way. Since rotation is not allowed, all these parts must appear in the chosen option at the same slant and in the same relative position. In option (c) the upper arc supplies the curved top, the bottom edge of the inner square supplies the straight base, and the two long slanting lines cross exactly as in the question figure. The piece of the arc caught between the two slanting lines is shorter than the base, which is precisely the shape the question figure shows. In option (a) the base is missing and the curves bend the other way, option (b) has no curve at all above the crossing point, and in option (d) the arcs sag downwards instead of arching upwards. Hence the question figure is hidden in option (c).

Embedded figure The question figure is made of only three straight lines: two of them are parallel and slope upwards to the right, while the third slopes downwards to the right and cuts across both of them. Since rotation is not allowed, the correct option must contain that same pair of parallel lines with one line crossing them at the same slant. Option (c) shows exactly this: two parallel chords running from the lower left to the upper right, and one chord running from the upper left to the lower right that cuts both of them. The two points of crossing lie one below the other, just as in the question figure, so the three lines sit together in the same way. Option (a) has only vertical and horizontal lines, option (b) replaces one of the parallel lines by a short stroke placed elsewhere, and option (d) has a vertical line but no pair of parallel slanting lines. Hence the question figure is embedded in option (c).

Counting triangles Name the parts first: ABCD is the square, O is the point where its two diagonals cross, M is the mid-point of AB, N is the mid-point of BC, and the horizontal line from O runs through N to the outside point P, which is joined to the points E and F of BC. Smallest triangles inside the square: AOM, MOB, BON, NOC, COD and DOA — that is 6. Two of the smallest ones join in pairs to give AOB (from AOM and MOB) and BOC (from BON and NOC) — 2 more. Each diagonal cuts the square into two halves, and every half is a triangle: ABC, ABD, BCD and ACD — 4 more. Outside the square the point P together with E, N and F gives PEN, PNF and PEF — 3 more. Total number of triangles = 6+2+4+3 = 15.

Counting triangles Name the points: ABCD is the rectangle, P the apex above it, E and F the mid-points of the two vertical sides, G the mid-point of the base and O the centre; PE, PG and PF cut the top edge at H, I and J. Above the rectangle the three lines from P give 3 triangles: PHI, PIJ and PHJ. One triangle is cut off at each corner of the rectangle: AHE, BJF, DEG and CFG — 4 triangles. The horizontal EF together with the apex gives PEO, POF and PEF — 3 triangles. The same horizontal together with G gives EOG, OFG and EFG — 3 triangles. Finally the two halves of the rhombus, PEG and PFG, add 2 more. Total number of triangles =3+4+3+3+2=15.

Counting triangles Name the points: ABCD is the upper rectangle with M the mid-point of AB and O the mid-point of DC, and DCE is the triangle hanging below it. AO produced meets CE at G, while MC crosses BO at P. Left half of the rectangle: AMO and ADO — 2 triangles. The right half is a square with both of its diagonals drawn, which always gives 8 triangles: MPB, BPC, CPO, OPM, MBC, BCO, COM and OMB. The two halves of the rectangle join to give one more, ABO. Below the horizontal: DOE on the left and COG, GOE, COE on the right — 4 triangles. The big lower triangle DCE and the slanting triangle MCE complete the list — 2 more. Total number of triangles =2+8+1+4+2=17.

Linear seating arrangement A sits next to both D and F, and F sits next to both E and A, so the four of them form one block: E – F – A – D. B is second from the right end, i.e. B occupies seat 5 of the six seats. The block of four must therefore take seats 1 to 4, and C is left with seat 6. D is not at either end, so the block cannot be read as D – A – F – E (that would put D in seat 1); it must be E – F – A – D, placing D in seat 4. E, F and A sit to the left of D, so 3 persons are sitting towards the left of D.

Statement and conclusion A conclusion may use only what the statement itself asserts; nothing may be brought in from outside. The e-mail asserts exactly one fact — the class timings will change from next week — and the second sentence only adds that every student has read it. So the only thing that certainly follows is that the timings of the classes are changing, which is option (b). A closure of the school, a vacation and a cancelled examination are fresh ideas that the e-mail never mentions, so none of them can be concluded.

Direction sense — turns A right turn is 90° clockwise and a left turn is 90° anticlockwise; going clockwise the order is north → east → south → west → north. R starts facing south. Take the turns of option (d): the first left turns him from south to east, the second left from east to north, and the right from north back to east. He finally faces east, so (d) is the required set of turns. Option (a) ends facing west (south → west → north → west), option (b) also ends facing west (south → east → south → west) and option (c) ends facing south.

Venn diagram — class relation The given diagram shows one class in the middle that overlaps each of the two outer classes, while the two outer classes themselves have no member in common. Option (c) fits this pattern: a principal can be a man or a woman, so 'Principal' overlaps both 'Man' and 'Woman'; but no man is a woman, so 'Man' and 'Woman' remain completely separate. In option (a) water, salt and soda are three different substances, so no class shares a member with another; in option (b) a tool is neither a man nor a teacher; in option (d) a groom is always male, so 'Groom' can never overlap 'Girl'. Hence the given Venn diagram best represents the classes of option (c).

Odd figure pair In every pair compare the four corners of the left square with the four corners of the right square. In (a) A, X, Y and 4 each travel to the diagonally opposite corner; the same happens to 2, 3, 0 and 7 in (b) and to the black square, M, P and the black circle in (d). So in (a), (b) and (d) the whole arrangement is turned through 180°. In (c) the arrow stays in the top-left corner and N stays in the bottom-right corner; only the circle and B exchange places, which is a reflection in the main diagonal, not a 180° turn. Pair (c) therefore obeys a different rule and is the odd one out.

Figure series Three things change from box to box: the black square, the small dot and the shaded half of the little circle. The black square runs through the cells in the order (1,1), (1,2), (2,1), (2,2), so the fifth figure must have it at (3,1). The small dot runs the same way but upward from the bottom: (4,2), (4,1), (3,2), (3,1), so next it stands at (2,2). The shaded half of the circle turns 90° clockwise each time — top, right, bottom, left — so the top half is shaded again, and the circle keeps changing sides, so it returns to the left of the box. Only option (a) shows the circle on the left with its top half shaded, the black square at (3,1) and the dot at (2,2). In (b) the lower half of the circle is shaded, in (c) the dot is at (3,2) and in (d) the left half of the circle is shaded, so all three are rejected.

Coding — number code Give every letter its place value in the alphabet: $A=1,B=2,\dots ,Z=26$. Add the place values of all the letters and then add half the place value of the first letter. BOOK: 2+15+15+11=43 and half of B is $\frac{2}{2}=1$, so the code is 43+1=44. PAGE: 16+1+7+5=29 and half of P is $\frac{16}{2}=8$, so the code is 29+8=37. PAPER: 16+1+16+5+18=56 and half of P is 8. Code for PAPER =56+8=64.

Word formation To write CLEAR we need the letters C, L, E, A and R, each at least once. CLEANER has C, L, E, A, N, E, R — all five letters are available. CALENDAR has C, A, L, E, N, D, A, R — all five letters are available. RECALL has R, E, C, A, L, L — all five letters are available. CANDLE has only C, A, N, D, L and E; there is no R in it, so CLEAR cannot be formed from CANDLE.